CP You are exploring a newly discovered planet. The radius of the planet is 7.20 X 107 m. You suspend a lead weight from the lower end of a light string that is 4.00 m long and has mass 0.0280 kg. You measure that it takes 0.0685 s for a transverse pulse to travel from the lower end to the upper end of the string. On the earth, for the same string and lead weight, it takes 0.0390 s for a transverse pulse to travel the length of the string. The weight of the string is small enough that you ignore its effect on the tension in the string. Assuming that the mass of the planet is distributed with spherical symmetry, what is its mass?

Solution 60P T The speed of a transverse wave in a string is given by v = m .....(, where T is the tension in the string and m is the mass per unit length of the string. Given, length of the string L = 4.00 m Mass of the string M = 0.0280 kg Therefore, mass per unit length of the string m = M/L = 0.0280/4.00 kg/m = 0.007 kg/m Time taken by the pulse to move from the lower to the upper end of the string = 0.0685 s Therefore, speed of this pulse in the planet vp= 4.00/0.0685 m/s = 58.4 m/s Substituting these values in equation (1), Tp 58.4 = 0.007 T p = (58.4) × 0.007 N = 23.87 N …..(2) Now, the speed of the pulse on earth is v e 4.00/0.0390 m/s = 102.6 m/s 102.6 = T e 0.007 T e = 73.7 N …..(3) This tension force will be equal to the earth’s gravitational force which can be expressed as, GM Me T e = R e2 73.7 = GM m2 …..(4) R e GM p Similarly, T p= 2 Rp 23.87 = GM p …..(5) R p2 Now, dividing equation (4) by (5), R p 2 M e 3.08 = ( R e) × M p …..(6) 7 Given, radius of the planet R = p.20 × 10 m Radius of earth R = 6.4 × 10 m 6 e 24 Mass of earth M = 5e97 × 10 kg We have to calculate the mass of the planet M . p From equation (6), 7 2 5.97×1024kg 3.08 = ( 7.20×106) × 6.4×10 M p 26 M p= 2.45 × 10 kg 26 So, the approximate mass of the planet is 2.45 × 10 kg .